How to Build a Supercapacitor Dragster on a Budget?

AI Thread Summary
The discussion centers on a project involving students building a supercapacitor-based dragster for a short race. The proposed design includes a lightweight structure made from wood and powered by a small DC motor (Gikfun DC 3V-6V Motor 2000 RPM) and supercapacitors. The calculations indicate that a single 2.7V, 10F supercapacitor could theoretically provide enough energy for the race, but there are concerns about the supercapacitor's ability to deliver the necessary current for high-energy applications. Participants emphasize the importance of checking the supercapacitor's datasheet for maximum output current and motor efficiency, as these factors significantly affect energy requirements. Additionally, there is a request for recommendations on commercially available devices to charge supercapacitors efficiently, as building a DIY charger may be too time-consuming for students. Overall, the discussion highlights the need for careful consideration of component specifications and energy calculations to ensure the project's success.
Fred404
Messages
2
Reaction score
3
Hi all!

I’m currently working on a project for which students will have to build a supercapacitor based dragster.

I saw here and there some videos on how it works but it’s too general to give me hints about components choice. So here I am with my questions, hoping that I’ll get enough answers to go forward with this project ;)

So, here it is:

I want students to build a small dragster (basically very light, made of wood sticks and such, not a big thing!) to organize a contest (drag race) on very small distance (let say about 3 meters max).

As for any school project, the amount of money is limited! No way to build a drag for 50$ each unfortunately.

Here are the components I was thinking about:

A small dc motor Gikfun DC 3V-6V Motor 2000 RPM

Here are specs:

EK2153.jpg


I was planning to feed it up with supercapacitors only. I was thinking of supercapacitor 2.7V, 10F to build a power pack in parallel if needed.

Here is what I’ve done so far (if I’m not wrong):

Energy stored in the supercapacitor 2.7V/10F

E=1/2.C.U2

E = (1/2)* 10F * 2.722V

E= 36.45 JoulesResistance of the DC motor:

R = U/I

At Maximum Efficiency – according to specs – we have 0.94A for 3V

R = 3V/0.9A = 3.34 ohm
Time of charge tho (sorry, don’t know the translation in English) for this setting:

Tho = R . C = 3.34 ohm x 10F = 33.4 s
So, this means that if tho = 33.4s, after 33.4s, there will only 37% energy left in capacitor.

After 5x tho, the capacitor will be empty.
Energy required for the race:

This is where I think, calculation may be inaccurate cause it depends on the weight of car (the heavier, the more energy needed)

Let say the car is about 150g (0.15 Kg)

F = m x g
F = 0.15 x 9.81 = 1.47 N

For a 3 meters race:

W = F x d
W = 1.47 x 3 = 4.41 J

For the motor, at Max Efficiency, we have 0.94A at 3V (even though the supercapacitor will only deliver 2.7V) for a 5 sec race

P = UI and E = Pt
P = 3 x 0.94 = 2.82W
E = 2.82 x 5 = 14.1 Joules

So, all "energy needed together", this represents an overall total of 4.41 + 14.1 = 18.51 J

According to that, this means that 1 supercapacitor would be enough
But just to have "backup", I can add an extra capacitor in parallel, which should give me twice energy...

So, am I totaly out of field? Does it make sens?

Also, does someone knows if I can buy a device to charge supercapacitors? I've found a diy to build one but it may take too much time for students to build it! So I was wondering if I could find a device that would do the job so that would just have to go to a station to charge their drag!

Thanks for any inputs ;)

Fred
 
Science news on Phys.org
I haven't used supercaps much, but my impression is that they are for low output currents (like real time clock backup), not for high energy applications. I could be wrong about that though. Have you looked in detail at the supercap datasheet to see what it shows for maximum output current?
 
Fred404 said:
F = m x g
No. The force will be the wheel torque divided by the wheel radius, but can never exceed the tire friction force which is the tire-road friction coefficient times the vertical force acting on the driven wheel. (More info here and here.)

Fred404 said:
So, all "energy needed together", this represents an overall total of 4.41 + 14.1 = 18.51 J
No. The energy needed will be the calculated output energy (4.41 J in the example) divided by the motor efficiency (43.8 % from the chart given), thus 4.41 / 0.438 = 10.14 J.
 
Thanks for those inputs: I'll check closer for both!
I haven't checked the datasheet for capacitors since I didn't ordered them yet and just check what is avail!
 
https://www.aapt.org/Conferences/ lists the next set of conferences 2026 Winter Meeting - January 17 - 19, Las Vegas, Nevada 2026 Summer Meeting - July 18 - 22, Pasadena, California 2027 Winter Meeting - January 9 - 12, New Orleans, Louisiana 2027 Summer Meeting - July 31 - August 4, Washington, DC I won't be attending the 2026 Winter Meeting in Las Vegas... For me, it's too close to the start of the semester. https://www.aapt.org/Conferences/wm2026/index.cfm...
Back
Top